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 2 years ago
Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:
 2 years ago
Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:

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Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0First Differentiate the function to find the slope.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0thank you but i dont know process. i am not sure about calculation, particulary differentiation

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0I mean i know the process

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0You know the process, but you need assistance with the differentiation portion?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, when it comes to both chain rule and products. I get really confused

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 2(x^2+y^2)^2=25(x^2−y^2)\]So we'll be differentiating this `implicitly`, meaning, `as is`. No solving for y ahead of time. When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable. \(\large y'(3,1)=m\). This is the m in our equation \(\large y=mx+b\). Hmm so let's see...

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0It appears we have no product rule which is nice.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0We should be able to get a numerical value for m, no x's or y's in our m. I'm not really sure what that Y7 is.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Taking the derivative of the left side gives us, `taking the derivative of the outermost function first, the square on the brackets`:\[\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}\]The blue is due to the chain rule. Understand that part?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}\]So we get something like this on the left side.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0i lied, I know product rule, power rule etc, but i dont know what the chain rule is

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0mmmm i always have trouble explaining that one. It's very unintuitive.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0could you give me example of chain rule problem maybe?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0The definition says this,\[\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)\]When you take the derivative of this, a function within a function, you ~Differentiate the outer function, f in this case. ~Then you make a copy of the inner function, and multiply by it's derivative. Here is a silly example. \[\large \frac{d}{dx}\left(x^2\right)^{4}\] In this example we're thinking of \(\large g(x)=x^2\). And \(\large f(x)=x^4\). If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), \(\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4\) will become, \(\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4\) which is \(\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4\). Applying the chain rule gives us,\[\large f'(g(x))=4\left(g(x)\right)^3 g'(x)\]See how the rule tells us to make a copy of the inner function and take it's derivative?\[\large f'(g(x))=4\left(x^2\right)^3 (x^2)'\]\[\large f'(g(x))=4\left(x^2\right)^3 (2x)\]

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Oo.. I see, Thank you. It was very clear! really cool :)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0If you were to simplify this down, you would get,\[\large f'(g(x))=8x^7\] Which we know is correct, because we could have easily approached this example without using the chain rule.\[\large y=(x^2)^4=x^8\]And we know the derivative will match what we got from the chain rule example.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0no, you explain well. i always get confused and could not tell wether i should use chain or product.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0So we applied the derivative of the left side,\[\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'\] right? :O

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0haha thats even more clear haha def, taking note

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0could we back to the first question?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0This one? :o \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0I see the chain rule now, (x^2+y^2)^2?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 2(x^2+y^2)\]That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.\[\large 2(x^2+y^2)(x^2+y^2)'\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 2(x^2+y^2)(2x+2y)\]So we should get somethingggggggg like this, yes?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0:O why two in the fron of the bracket? I understand the (2x+2y) tho

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0You applied the power rule to the outer function \(\large \left(\qquad \right)^2\)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0power rule and chain double? :O

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Power rule the outside, make a copy of the inside, take derivative of copy.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})\]See this blue term? We took the derivative of a variable that is not x, namely y, so a y' popped out.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0mmmmmm.... Ok. I think :/

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0so whenever its chain rule i put y' in the end?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0No. Let's not think of that portion as chain rule, I think that's just going to confuse you. Whenever we take the derivative of y, we will multiply by y'. We don't throw it on the end, we always attach it to the y that we're taking the derivative of.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0right, Ok. i will remember that by my heart.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Try taking the derivative of the right side, tell me what you get.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0close, but you took the derivative of a y, you need to multiply that part by y'

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0its still funny how i have to put y' in the end. is it for implicit derviation?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0yes. If we had an equation like this,\[\large y=x^2\]The derivative gives us,\[\large y'=2x\]Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?\[\large y^2=x^2\]Taking the derivative,\[\large 2y=2x\] Hmm we don't have a derivative term do we?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0so 2(x^2+y~2)(2x+2y)=25(2x2yy')

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.02(x^2+y^2)^2=25(x^2−y^2)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0See that 2 at the front? That would give us an extra factor of 2 in our derivative.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0the question equation is 2(x^2+y^2)^2=25(x^2y^2)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0\[4(x^2+y^2)(2x+2y)=25(2x2yy')\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x2yy')\]Woops you missed a y'.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0ops! haha thats derivative of the first equation?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0yes. We took the derivative of y on each side, so each y will get a y'

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate \(\large (3,1\;)\) and THEN solve for y'.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0plug both number into the equation?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0yes, 3 for all the x's. 1 in for all the y's.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0and don't plug anything in for y'. That's what we'll be solving for.
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